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Consider free scalar field in two dimensions with the standard action written in complex coordinates $S=\int d^2z\, \partial \phi\bar{\partial}\phi$. The two-point correlation function is known to be $$\left\langle\phi(z,\bar{z})\phi(w,\bar{w})\right\rangle = \alpha\ln|z-w|^2$$ here $\alpha$ is some unimportant constant.

On the other hand, one expects from the conformal symmetry that the correlator must be invariant under any holomorphic variable charge. Apparentely, it is not.

I am aware that the conformal symmetry is in some sense anomalous -- the symmetry algebra of the QFT is not the Witt algebra, but its central extension, the Virasoro algebra.

The question is whether this anomaly is seen right at the simplest example of the two-point correlation function for free scalar field? Or am I confused about something?

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    $\begingroup$ Correlators are covariant under symmetry transformations, not invariant. $\endgroup$ – Prahar Mar 27 '16 at 5:42
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    $\begingroup$ PS - Free scalar fields are not operators in the CFT. They are used to construct well-defined operators, such $\partial \phi$ and ${\bar \partial} \phi$, but they themselves are not operators. $\endgroup$ – Prahar Mar 27 '16 at 13:39
  • $\begingroup$ Could you elaborate on why aren't they operators? I know that they are not primary, but are you saying that their action is not well-defined on the Hilbert space? $\endgroup$ – Weather Report Mar 28 '16 at 18:53
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    $\begingroup$ Their correlation functions are IR divergent. Thus the states they would create under the state operator map are non-normalizable. $\endgroup$ – Prahar Mar 28 '16 at 18:55
  • $\begingroup$ Interesting point. However, this seems to be a feature of 2d, not of the conformal invariance (if am I not mistaken, for the massive scalar field the propogator is still IR-divergent)? $\endgroup$ – Weather Report Mar 28 '16 at 20:36
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The correlator in question is not invariant even under scaling transformations, and the $sl_2\mathbb{C}\times sl_2\mathbb{C}$ subalgebra of global conformal transformations (generated by $L_{\pm 1},L_0$ and their anti-holomorphic counterparts), which contains the scaling, of Virasoro algebra is always anomaly-free in the sense that you mention. (You can check that the central term in commutation relations vanishes for $n=\pm 1,0$.)

There is conformal anomaly on curved manifolds, but the correlator you write is in flat space.

So the answer is no, the issue you rise is not related to conformal anomaly in any obvious way.

You can think of this as that the correlation functions of free scalar require a regularization, which cannot be removed completely and you get a scale dependence. It is somewhat similar to an anomaly, I guess, except that this correlator is most likely scheme dependent. In CFT language, the field $\phi$ itself is not a primary field, and its correlators are not required to have the standard transformation properties.

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  • $\begingroup$ I thought that the very definition of the anomaly is breaking of the classical symmetry at the quantum level. From the classical theory we expect on symmetry grounds that the scalar field is the primary operator of dimensions $\Delta=\bar{\Delta}=0$. This turns out false. Why should not this be called an anomaly? Thanks to your answer I see that it is not directly related to the central charge. However, this makes things even more puzzling to me. I would like to see some further clarifications, if possible. Thanks! $\endgroup$ – Weather Report Mar 28 '16 at 19:12

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